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    (Original post by justinawe)
    Hmm?

    You understand that m at Q is -\dfrac{1}{2} ?

    You understand this means that \dfrac{dy}{dx} at Q is -\dfrac{1}{2} ?

    You understand that y = \dfrac{1}{4}x^2 \Rightarrow \dfrac{dy}{dx} = \dfrac{1}{2}x ?

    and thus \dfrac{1}{2}x = -\dfrac{1}{2} ?
    Ohhhhh..... OK... !

    its just a revers - ie thing

    I got it! /// Wow!! - Your maths skills are comparable to a DON!!

    i would rep but im out
    thank you!

    ryan
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    (Original post by purplemind)
    It would be really nice.
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    (Original post by tigerz)
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    Thank you.
    The attached document looks quite useful.
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    (Original post by purplemind)
    Thank you.
    The attached document looks quite useful.
    Haha no problem
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    (Original post by tigerz)
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    Daymnn guuurl loving those colorful notes
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    (Original post by ryanb97)
    Ohhhhh..... OK... !

    its just a revers - ie thing

    I got it! /// Wow!! - Your maths skills are comparable to a DON!!

    i would rep but im out
    thank you!

    ryan
    Oh come now, I'm just more experienced you're in year 10 or 11 iirc...
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    (Original post by tigerz)
    Haha no problem
    Anyway, I'm being a bit thick now, but seriosly, I don't know how to go about doing it.
    I am given P(A)=3/4
    P(B|A)=1/5
    P(B'|A')=4/7
    They want me to find P(A and B)- that's easy, I did it. But then they want P(B). And it confuses me a bit.
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    c2 friday aaaaarghhh
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    (Original post by Boy_wonder_95)
    Daymnn guuurl loving those colorful notes
    I didn't realise you were american :hmmm:
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    (Original post by purplemind)
    They want me to find P(A and B)- that's easy, I did it. But then they want P(B). And it confuses me a bit.
    Surely the \displaystyle P(B) = P(B|A) + P(B|A') ?

    EDIT: I meant \displaystyle P(B) = P(A \cap B) + P(A' \cap B)
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    (Original post by Scientific Eye)
    Surely the \displaystyle P(B) = P(B|A) + P(B|A') ?
    But why?
    Even if they are not mutally exclusive?
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    (Original post by justinawe)
    Oh come now, I'm just more experienced you're in year 10 or 11 iirc...
    year 11 .... but i must say i havent been in a proper maths lesson in 4 years :cry2:

    and i wont be next year...:cry:

    lol

    ryan
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    (Original post by justinawe)
    I didn't realise you were american :hmmm:
    Haha, is that a bad thing? :hmmm:
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    (Original post by Boy_wonder_95)
    Daymnn guuurl loving those colorful notes
    (Original post by justinawe)
    I didn't realise you were american :hmmm:
    LOOL! IKR thank youuu, but I only just realised they have loads of spelling errors btw, we're literally meeting on every thread :laugh:
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    (Original post by Boy_wonder_95)
    Haha, is that a bad thing? :hmmm:
    No, of course not. Are you really, though?
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    (Original post by justinawe)
    No, of course not. Are you really, though?
    Nah , it's just that TSR has 'colourful' underlined in red which buggers me :rolleyes:
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    (Original post by purplemind)
    But why?
    We are only concerned with the outcomes where event \displaystyle B occurs.

    Look at the probability tree. You have event \displaystyle A and event \displaystyle B. We don't care if event \displaystyle A happens or not, only that event \displaystyle B does happen. Hence we find all the ends of the probability tree where event \displaystyle B is an outcome and then add them up.

    That is, if I'm not incorrect in my thinking.
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    (Original post by purplemind)
    Anyway, I'm being a bit thick now, but seriosly, I don't know how to go about doing it.
    I am given P(A)=3/4
    P(B|A)=1/5
    P(B'|A')=4/7
    They want me to find P(A and B)- that's easy, I did it. But then they want P(B). And it confuses me a bit.
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    Use these two formulae (which are in the formulae booklet):


    P(A \cap B) = P(A)P(B | A)

    P(A \cup B) = P(A) + P(B) - P(A \cap B)


    Find P(A \cap B) using the first formula, then sub it into the second one.
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    (Original post by tigerz)
    LOOL! IKR, but I only just realised they have loads of spelling errors btw, we're literally meeting on every thread :laugh:
    Still they make my S1 notes look rusty in comparison , who me or justin?
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    (Original post by Scientific Eye)
    We are only concerned with the outcomes where event \displaystyle B occurs.

    Look at the probability tree. You have event \displaystyle A and event \displaystyle B. We don't care if event \displaystyle A happens or not, only that event \displaystyle B does happen. Hence we find all the ends of the probability tree where event \displaystyle B is an outcome and then add them up.

    That is, if I'm not incorrect in my thinking.
    Ok, so P(B|A)=1/5
    P(B|A')=3/7
    I add them and get P(B)=22/35
    My textbook says it should be 9/35
    So I don't know if it is right and the answer in the textbook is wrong or the other way round. :confused:

    (Original post by justinawe)
    Use these to formulae (which are in the formulae booklet):


    P(A \cap B) = P(A)P(B | A)

    P(A \cup B) = P(A) + P(B) - P(A \cap B)


    Find P(A \cap B) using the first formula, then sub it into the second one.
    I still don't know how to find P(A U B) from what I am given. :<
 
 
 
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